Question

A small body of mass $m=10\text{kg}$ hangs at one end of a string of length $l$, the other end of which is fixed. It is given horizontal velocity $u$ at its lowest position. The string just becomes slack, when it covers an angle ${120}^{\circ }$ with the initial position. Find the tension in the string at the point of projection. Take $g=10{\text{m/s}}^2$

• A. $900\text{N}$
• B. $50\text{N}$
• C. $5\text{N}$
• D. $450\text{N}$

Answer 1

The correct option is D $450\text{N}$

Initially, FBD of the body is:


$T-mg=\frac{m{u}^2}{l}$ ... (1)
where $u=$ initial velocity given to body

Finally, the string will make angle ${180}^{\circ }-{120}^{\circ }={60}^{\circ }$ with the vertical.


$mg\cos {60}^{\circ }+T=\frac{m{v}^2}{l}$
(It slack at $\theta ={60}^{\circ }$. Hence $T=0$)
$\Rightarrow \frac{m{v}^2}{l}=\frac{mg}{2}$ ... (2)

By work energy theorem (taking P.E at bottom = 0)
$0+\frac{1}{2}m{u}^2=mg(l+l\cos {60}^{\circ })+\frac{m{v}^2}{2}$
$\Rightarrow \frac{m{u}^2}{2}=\frac{3mgl}{2}+\frac{m{v}^2}{2}$
$\Rightarrow m{u}^2-m{v}^2=3lmg$... (3)

Divide by $l$
$\frac{m{u}^2}{l}-\frac{m{v}^2}{l}=3mg$
From eq. (1) and (2)
$T-mg-\frac{mg}{2}=3mg$
$\Rightarrow T=3mg+\frac{3mg}{2}=\frac{9mg}{2}$
$=\frac{9\times 10\times 10}{2}$
$\therefore$ tension at bottommost point $T=450\text{N}$